Suppose we have a surface $S$ defined by the transformation $T$ for $\dfrac{-\pi}{2} < u < \dfrac{\pi}{2}$ and $0 < v < \pi$. $T(u, v) = (u, \cos(u), \sin(v))$ What is the surface area of $S$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin(v) \cos(u) \, du \, dv$ (Choice B) B $ \int_0^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} | \cos(v) \cos(u) | \, du \, dv$ (Choice C) C $ \int_0^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} | \cos(v) | \sqrt{\sin^2(u) + 1} \, du \, dv$ (Choice D) D $ \int_0^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \cos(v) \sqrt{\cos^2(u) + 1} \, du \, dv$
Solution: Assume we have a surface $S$ parameterized by a transformation $T$. If we want to find the surface integral over $S$ of a function $f$, we can use the formula below to convert it into a familiar double integral. $ \iint_S f(T(u, v)) | T_u \times T_v | \, du \, dv$ Finding surface area using a surface integral means using $f(x, y, z) = 1$. In effect, we are saying that we only care about the scaling factor caused by the area element. Therefore: $A = \int_0^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} |T_u \times T_v| \, du \, dv$ Now we need to find the magnitude of the area element. $|T_u \times T_v| = | \cos(v) | \sqrt{\sin^2(u) + 1}$ [Calculation] In conclusion, the surface area of $S$ is: $ \int_0^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} | \cos(v) | \sqrt{\sin^2(u) + 1} \, du \, dv$